Question: Simplify the following expression: $y = \dfrac{-7x^2+34x+5}{x - 5}$
First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-7)}{(5)} &=& -35 \\ {a} + {b} &=& &=& {34} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-35$ and add them together. Remember, since $-35$ is negative, one of the factors must be negative. The factors that add up to ${34}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-1}$ and ${b}$ is ${35}$ $ \begin{eqnarray} {ab} &=& ({-1})({35}) &=& -35 \\ {a} + {b} &=& {-1} + {35} &=& 34 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-7}x^2 {-1}x) + ({35}x +{5}) $ Factor out the common factors: $ x(-7x - 1) - 5(-7x - 1)$ Now factor out $(-7x - 1)$ $ (-7x - 1)(x - 5)$ The original expression can therefore be written: $ \dfrac{(-7x - 1)(x - 5)}{x - 5}$ We are dividing by $x - 5$ , so $x - 5 \neq 0$ Therefore, $x \neq 5$ This leaves us with $-7x - 1; x \neq 5$.